how to find a primitive of $(s^2+2s+1)e^{\frac12 s^2}$?

Question

in a certain problem I had to differentiate the following function :

$$\xi (s) = se^{\frac12 s^2}+ 2e^{\frac12 s^2}$$

I got $$\xi '(s) = (s^2+2s+1)e^{\frac12 s^2}$$

but it seems much harder to go backwards

how do you integrate this kind of functions ?

Answer 1

Let $$I=\int s(s+1)e^{\frac 12s^2}ds$$ and $$J=\int (s+1)e^{\frac 12s^2}ds$$

Do $I$ by parts with $u=s+1$ and get $$I=(s+1)e^{\frac 12s^2}-\int e^{\frac 12s^2}ds$$

Do $J$ by splitting it as $$J=\int se^{\frac 12s^2}+\int e^{\frac 12s^2}ds$$ $$=e^{\frac 12s^2}+\int e^{\frac 12s^2}ds$$

Then $$I+J=(s+2)e^{\frac 12s^2}$$

Answer 2

You use integration by parts. In general:

\begin{align} \ \int x^ne^{\frac 12 x^2}dx & =\int x^{n-1} \cdot xe^{\frac 12 x^2}dx \\ \ & = x^{n-1}e^{\frac 12 x^2}-(n-1)\int x^{n-2} \cdot e^{\frac 12 x^2}dx \\ \ & = (x^{n-1}-(n-1)x^{n-3})e^{\frac 12 x^2}+(n-1)(n-3)\int x^{n-4} \cdot e^{\frac 12 x^2}dx \end{align}

and so on.

For odd $n$, you will eventually get down to $\int xe^{\frac 12 x^2}dx$ which integrates to just $e^{\frac 12 x^2}$.

As for even $n$, you get $\int e^{\frac 12 x^2}dx$ which is the error function.

Answer 3

Hint: Substitute $$x=t-1$$ so you will get $$dx=dt$$ and your integral will be $$\int t^2e^{\frac{1}{2}(t-1)^2}dt$$ and use Integration by parts.